1) The sum of the squares of the two natural no is 52 ,If the first no is 8 less than twice the second no ,find the nos.2)Two water taps together can fill a tank in hrs .The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank respectively .Find the time in which each tap can separately fill a tank.Kindly give both answers

1) The sum of the squares of the two natural no is 52 ,If the first no

1.	1) The sum of the squares of the two natural no is 52 ,If the first no is 8 less than twice the second no ,find the nos.2)Two water taps together can fill a tank in hrs .The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank respectively .Find the time in which each tap can separately fill a tank.Kindly give both answers
Answer» HEY user Here is your answer :- Q.1 Let the one number be ' X ' . So other number will be ' 2x - 8 ' By the condition given in the question ${x}^{2} + {(2x - 8)}^{2} = 52 \\ \\ {x }^{2} + {(2x)}^{2} - 2(2x)(8) + {8}^{2} = 52 \\ \\ {x}^{2} + 4 {x}^{2} - 32x + 64 = 52 \\ \\ 5 {x}^{2} - 32x + 64 - 52 = 0 \\ \\ 5 {x}^{2} - 32x + 12 = 0 \\ \\ 5 {x}^{2} - 30x - 2x + 12 = 0 \\ \\ 5x(x - 6) - 2(x - 6) = 0 \\ \\ (x - 6)(5x - 2) = 0 \\ \\ (x - 6)= 0 \: \: \: \: \: \: or \: \: \: \: \: \: 5x - 2 = 0 \\ \\ x = 6 \: \: \: \: \: \: or \: \: \: \: \: \: x = \frac{5}{2} \\ \\ x \: is \: a \: natural\: number \: \\ \\ hence \: x = 6 \\ \\ other \: number \: = 2x - 8 \\ \\ = 2(6) - 8 \\ \\ = 12 - 8 \\ \\ = 4 \\ \\ so \: the \: numbers \: are \: 6 \: and \: 4$ _____________×○×___________ Q.2 Let the time TAKEN by tap of larger diameter to fill the tank be " x hours " . The tap can fill 1/x part of tank in a HOUR So the time taken by tap of smaller diameter is " x + 10 hours " . The tap can fill 1/x+ 10 part of RANK in a hour . $both \: the \: taps \: can \: fill \: the \: tank \: in \: 9 \times \frac{3}{8} = \frac{75}{8} \\ \\ so \: they \: can \: fill \: \frac{8}{75} \: part \: of \: tank \: \\ \\ so \: we \: get \: equation \: \\ \\ \frac{1}{x} + \frac{1}{x + 10} = \frac{8}{75} \\ \\ \frac{(x + 10) + x}{x(x + 10)} = \frac{8}{75 } \\ \\ \frac{2x + 10}{ {x}^{2} + 10x} = \frac{8}{75} \\ \\ 8( {x}^{2} + 10x) = 75(2x + 10) \\ \\ 8 {x}^{2} + 80x = 150x + 750 \\ \\ 8 {x}^{2} + 80x - 150x - 750 = 0 \\ \\ 8 {x}^{2} - 70x - 750 = 0 \\ \\ comparing \: the \: equation \: with \: a {x}^{2} + bx + c = 0 \\ \\ here \: a \: = 8 \\ b = - 70 \\ c = - 750 \\ \\ {b}^{2} - 4ac = {( - 70)}^{2} - 4(8)( - 750) \\ \\ = 4900 + 24000 \\ \\ = 28900 \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a } \\ \\ = \frac{ - ( - 70) + - \sqrt{28900} }{2 \times 8} \\ \\ = \frac{70 + - 170}{16} \\ \\ x = \frac{70 + 170}{16} \: \: \: \: or \: \: \: \: x = \frac{70 - 170}{16} \\ \\ x = \frac{240}{16} \: \: \: \: or \: \: \: \: x = \frac{ - 100 }{16} \\ \\ x = 15 \: \: \: \: or \: \: \: \: x = \frac{ - 25}{4} \\ \\ \: time \: cannot \: be \: negative \: \\ \\ x = 15 \\ \\ x + 10 = 15 + 10 = 25 \\ \\ so \: the \: taps \: with \: large \: and \: small \:diameter \: can \: fill \: the \: tank \: in \: 15 \: and \: 25 \: hours \: respectively \: indivitually$

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